This question was asked in my assignment of Smooth manifolds course and I am struct on this problem.
Show that $SO(3) =${$ A\in M(3,n) |A^T A=I , det A= 1$ } is a submanifold of $\mathbb{R}^3$.
I found a solution while searching this website: Show that $SO(3)$ in an embedded submanifold of $\mathbb{R}^{3 \times 3}$.
but this solution uses Regular Level Set Theorem which has not been covered in my course.
I tried by using the definition of Submanifold which is: I have to show that SO(3) is a k-dimensional submanifold of M(3,n) : If $P\in SO(3) $ P being a point, there is a chart $(U, \phi)$ containing P such that $\phi(U \cap SO(3,n))$ is an open subset of $\mathbb{R}^9$.
Here $U\subseteq SO(3)$ . But I am not able to move foreward from here.
Kindly help me!
Jean Gallier's notes: https://www.cis.upenn.edu/~jean/gma-v2-chap5.pdf See Page 55/ Question 5.4.(ii).
Let $V_3:S^3\rightarrow\Bbb{R}^{10}$ be the Veronese map $V_3(x,y,z,t)=(x^2,y^2,z^2,t^2,xy, xz, xt, yz, yt, zt)$. Then, since $V_3(-x,-y,-z,-t)=V_3(x,y,z,t)$ it induces an embedding $\phi: \Bbb{P}^3\rightarrow \Bbb{R}^{10}$ such that $V_3=\phi\circ\pi$ where $\pi: S^3\rightarrow\Bbb{P}^3$ is the quotient map. On the other hand, the image of $V_3$ is into the hyperplane $H\subset\Bbb{R}^{10}$ defined by the equation $x_1+x_2+x_3+x_4=1$. Since $H\cong\Bbb{R}^{9}$, we have an embedding $\psi: SO_3\cong\Bbb{P}^3\rightarrow \Bbb{R}^{9}$. But, I don't know how to make this map more explicit. Also, this is an algebraic geometric approach. I am trying to find an algebraic topologic approach which uses the canonical line bundle of the projective space.
We have also the obvious embedding $SO(3)\subset M(3,\Bbb{R})=\Bbb{R}^9$.
$SO(3)=\Bbb{P}^3$ can not be embedded in $\Bbb{R}^4$ but it embedds in $\Bbb{R}^5$. To do this embedd the projective plane $\Bbb{P}^2$ in $\Bbb{R}^4$, then choose a point $x_0\in\Bbb{R}^5$ which is not in $\Bbb{R}^4\subset\Bbb{R}^5$. Then, construct the cone with base $\Bbb{P}^2$ and apex $x_0$. This cone is $SO(3)$ (?).